LeetCode 169. Majority Element

Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.

You may assume that the array is non-empty and the majority element always exist in the array.


A rather straight forward method is to use hashmap. However, it is a waste of memory space.

This is actually a pretty famous algorithm called: MJRTY - A Fast Majority Vote Algorithm  by Robet S. Boyer.

#include <vector>
#include <iostream>
#include <climits>
using namespace std;

int majorityElement(vector<int>& nums) {
    int count = 0;
    int target = INT_MAX;
    int i = 0;
    while(i < nums.size()) {
        if(count == 0) {
            target = nums[i];
            count++;
        } else if(nums[i] == target){
            count++;
        } else {
            count--;
        }
        i++;
    }
    count = 0;
    for(i = 0; i < nums.size(); ++i) {
        if(nums[i] == target) count++;
    }
    if(count > nums.size() / 2) return target;
    else return -1; // remember to verify, it maybe has no majority. For example [1, 1, 2, 2]
}

int main(void) {
    vector<int> nums{1, 1, 2};
    int target = majorityElement(nums);
    cout << target << endl;
}



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